120 mL of pure phosphine is decomposed to produce vapours of phosphorus and `H_2`. The change in volume during reaction is :

Options

A. 60mL
B. 75mL
C. 80mL
D. 180mL

Explanation

To solve the problem of determining the change in volume when 120 mL of pure phosphine (PH₃) decomposes to produce vapors of phosphorus (P) and hydrogen (H₂), we can follow these steps:

### Step 1: Write the decomposition reaction of phosphine
The decomposition of phosphine can be represented by the following balanced chemical equation:

\[ 4 \, \text{PH}_3 \rightarrow \text{P} + 6 \, \text{H}_2 \]

### Step 2: Analyze the stoichiometry of the reaction
From the balanced equation, we see that:
- 4 moles (or volumes) of phosphine (PH₃) decompose to produce 1 mole (or volume) of phosphorus (P) and 6 moles (or volumes) of hydrogen (H₂).

### Step 3: Calculate the volume of hydrogen produced from the given volume of phosphine
Given that we start with 120 mL of phosphine (PH₃), we can calculate the volume of hydrogen produced using the stoichiometric ratios from the balanced equation.

Using the ratio from the balanced equation:
- 4 mL of PH₃ produces 6 mL of H₂.

Now, we can set up a proportion to find out how much hydrogen is produced from 120 mL of PH₃:

\[ \text{Volume of H}_2 = \left( \frac{6 \, \text{mL H}_2}{4 \, \text{mL PH}_3} \right) \times 120 \, \text{mL PH}_3 \]

Calculating this gives:

\[ \text{Volume of H}_2 = \frac{6}{4} \times 120 = 180 \, \text{mL} \]

### Step 4: Determine the change in volume during the reaction
The change in volume can be calculated by subtracting the initial volume of phosphine from the total volume of gases produced (which is the volume of hydrogen, since the volume of phosphorus is negligible).

\[ \text{Change in volume} = \text{Volume of H}_2 - \text{Initial volume of PH}_3 \]

Substituting the values we have:

\[ \text{Change in volume} = 180 \, \text{mL} - 120 \, \text{mL} = 60 \, \text{mL} \]

### Conclusion
Thus, the change in volume during the reaction is **60 mL**.

---